How do you find the vertex and intercepts for #y = x^2 - 9#?

1 Answer
May 30, 2018

Vertex: #(0,-9)#
#y"-intercept: " y=-9#
#x"-intercepts: " x=-3" and " x=+3#

Explanation:

The general vertex form is
#color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)bcolor(white)("xxx")#with vertex at #(color(red)a,color(blue)b)#

The given equation #y=x^2-9# can be written explicitly in vertex form as
#color(white)("XXX")y=color(green)1(x-color(red)0)^2+color(blue)(""(-9))color(white)("xxx")#with vertex at #(color(red)0,color(blue)(""(-9)))#

The #y#-intercept is simply the value of #y# when #x=0#.
Substituting #0# for #x# in the given equation:
#color(white)("XXX")y=0^2-9rArry=-9#

Similarly, the #x#-intercepts are the values of #x# which solve the equation when #y=0#.
Substituting #0# for #y# in the given equation:
#color(white)("XXX")0=x^2-9#
#color(white)("XXX")rarr x^2=9#
#color(white)("XXX")rarr x=+-3#